How complicated can a finite double complex over a field be?

Question

A finite complex over a field $k$ is pretty simple: it's the direct sum of its homology with a split-exact complex. How complicated can a finite double complex be? Does it make a difference if $k$ is algebraically closed?

It seems as though quiver representation theory may be relevant here, but since a double complex is a representation of a quiver with relations, I'm not really sure where to start looking. So my question is:

Question: Let $k$ be a field. How complicated is the category of finite-(total) dimensional double complexes over $k$?

Answer 1

It seems that Jonas Stelzig has answered your question completely, for the sake of applications to non-Kähler geometry. $\require{AMScd}$

I will quote here Theorem A, which enumerates all of the isomorphism classes of double complexes as direct sums of certain standard double complexes. Because it is easy to write down the set of maps between any two of the standard double complexes, I am hopeful this description is enough to determine any categorical facts you might wonder about.

For any bounded double complex over a field $K$, there exist unique cardinal numbers $\text{mult}_S(A)$ and a non-functorial isomorphism $A \cong \bigoplus_S S^{\oplus \text{mult}_S(A)}$, where $S$ runs over squares and zig-zags:

$$\begin{CD}K @>-1>>K\\@A1AA @A1AA\\K @>1>> K\end{CD}, \;\;\;K,\;\;\; K \xrightarrow{1} K, \;\;\; \begin{CD}K \\ @A1AA \\ K\end{CD}, \;\;\; \begin{CD}K \\ @A1AA \\ K @>1>> K\end{CD}, \;\;\; \cdots$$

That is, ignoring bigradings, $S$ consists of the square, the single-element term, and two zig-zags of each length $L > 0$ (depending on orientation of the 'initial' arrow); there are $\Bbb Z^2$ each of these generators, depending on the bigrading of the initial element.

Answer 2

The representation theory of those representations concentrated in the $n \times m$-grid are just representations of the tensor product of the radical square zero algbras $A_n$ and $A_m$ with underlying quiver a linear oriented line. For $n=m$, they have been also classified in https://arxiv.org/pdf/1703.08377.pdf (see proposition 2)and this then also gives the classificaiton for arbitrary $n$ and $m$ when looking at the indecomposables which lie only in the $n \times m$ grid for $m \leq n$.

Here another proof using some theorems. Let $C=A_n \otimes_K A_m$. By Theorem 4.5. a) in chapter IX. of the book "Elements of the representation theory of associative algebras" by Assem, Simson and Skowronski the algebra $C$ has a postprojective component. Thus one just have to show that for every indecomposable projective $C$-module $P$ one has $\tau^{-l}(P)=0$ for $l$ large enough to conclude that there are only finitely many indecomposable modules and all are isomorphic to $\tau^{-r}(P)$ for some $r$ and some indecomposable projective module $P$. (starting from this one can get the Auslander-Reiten quiver of $C$ easily)

Seeing those complexes as tensor products also shows that in any dimension higher than 2, there are infinitely many indecomposable representations in general as already $A_2 \otimes_K A_2 \otimes_K A_2$ is representation-infinite.